SOLUTION.
The area of the entire region will be to find the area of the isosceles triangle and the semi-circle.
[tex]\begin{gathered} \text{Area of the triangle is } \\ =\frac{1}{2}\times a\times b\times\sin \theta \\ \text{Where } \\ \text{where a and b are two sides and }\theta=angle\text{ betwe}en\text{ the two sides } \end{gathered}[/tex]
Then
[tex]\begin{gathered} =\frac{1}{2}\times8\times8\times\sin 72.6 \\ =32\sin 72.6 \\ =30.54\text{unit}^2 \end{gathered}[/tex]
The area of the isosceles triangle is 30.354 square units
We need to obtain the diameter of the semi-circle using the cosine rule
We need to obtain the value of x,
[tex]\begin{gathered} x^2=8^2+8^2-2(8)(8)\cos 72,6 \\ x^2=128-128\cos 72.6 \\ x^2=128-38.27 \\ x^2=89.72 \end{gathered}[/tex]
Take square root of both sides, we have
[tex]x=\sqrt[]{89.72}=9.74[/tex]
Hence, the diameter is 9.74
Then the radius will be
[tex]r=\frac{9.74}{2}=4.87[/tex]
The area of the semi-circle is
[tex]\begin{gathered} =\frac{1}{2}\pi r^2 \\ =\frac{1}{2}\times3.14\times4.87^2 \\ =37.24\text{unit}^2 \end{gathered}[/tex]
The area of the semi circle is 37.42 square units
Therefore the area of the region will be
[tex]\begin{gathered} =\text{Area ot triangle + Area of semi circle } \\ =30.54+37.42 \\ =67.96uniits^2 \end{gathered}[/tex]
Thus
The area of the region is 67.96 units²
